Fleming's left-hand rule on charged particles

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moenste
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Homework Statement


Diagrams (a) to (c) show a magnetic field of flux density 0.2 T directed perpendicularly into the paper. In each of (a) and (b) a conductor of length 0.3 m is entirely within the field and is carrying a current of 4 A in the plane of the paper. In (c) an electron is moving in the plane of the paper at 2 * 106 m/s. Copy the diagrams and show the direction of the force in each case. Also find the magnitude of the forces. (Charge on the electron = 1.6 * 10-19 C.)

Images:
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Answers:
96706ca803d9.jpg

2. The attempt at a solution
F = BIL sin θ
F = BQv sin θ

(a-b) F = 0.2 * 4 * 0.3 * sin 90 = 0.24 N
(c) F = 0.2 * 1.6 * 10-19 * 2 * 106 * sin 90 = 6.4 * 10-14 N

The (a-b) graphs we get using the Fleming's left-hand rule. For (c), however, I get the force directed in the opposite direction (to north-east, not south-west). I think this is because (in constrast to (a-b), where there was current in each case and not particle) the particle is directed downwards and thus is negatively charged, so we need to change the middle finger from south-east to north-west. But I'm not sure about that. Any ideas on (c) graph please?
 
Last edited:
on Phys.org
haruspex said:
If an electron is moving down and to the right, which way is the current?
The opposite way, so upwards and left. And having this using the left-hand rule we get the correct answer. Right?